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I am not properly understanding this question. please help!

Most of the process on PV diagram have this kind of curvature

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But in the question above, its bit weird.

My thoughts:

Pressure, temperature and volume all increasing.

So It looks like combustion process than expansion.

I am unable to figure out answer for it. They all seems correct.

There is definitely net work (-ve may be but there is net work)

Not sure about Net heat output.

As expanding density might decrease but pressure is increasing so it might increase.

And enthalpy is more likely to increase as all the quantities are increasing.

Please clarify the above thoughts and explain the answer.

I searched a lot all over the internet. Did not found any process look similar to this!


1 Answer 1


The process is general in a closed system. The gas is ideal. Adding isotherms on the $p,V$ axes will show whether $T$ increases / decreases as $p$ and $V$ increase. Work is $-\int p_{ext}\ dV$. An assumption of reversibility or constant external pressure will be needed to determine whether 1) is / is not true. The internal energy of an ideal gas depends only on the temperature of the ideal gas. The use of $\Delta U = C_V \Delta T = q + w$ will tell whether 2) is / is not true. The system is closed, so the amount (moles) of gas is constant. Volume changes and density is the inverse of molar volume. The will address whether 3) is / is not true. The enthalpy of an ideal gas also depends only on its temperature. This will address whether 4) is / is not true.

  • $\begingroup$ Is this adiabatic process in question? or heat is being added to increase the temperature? $\endgroup$ Commented May 5, 2021 at 2:18
  • $\begingroup$ The question is answered by solving part 2. What did you learn when you did part 2? $\endgroup$ Commented May 5, 2021 at 14:51
  • $\begingroup$ In part 2 you said internal energy depends on temperature. In the question temperature is increasing. As output temperature is higher, we will get +ve change in internal energy Right? and about net heat output q = CvDel(T) - (-W). So both will have positive signs and there will be +ve heat output ? $\endgroup$ Commented May 10, 2021 at 12:59
  • $\begingroup$ Positive $q$ indicates an endothermic process. $\endgroup$ Commented May 11, 2021 at 18:23
  • $\begingroup$ Thanks for the answer. I was confused with First law. I was using ΔU = ΔQ + ΔW instead of ΔU = Q +W to visualize it making simple thing complicated. Let me sum up the answers 1) there is definately work done as volume is changing. 2) As there is work done and temperature change ( as T2 > T1) , there has to be ΔU so as Q. 3) Volume increasing so density will decrease. 4) Temperature also increased, so there will be enthalpy increase as well. Now I think I understood your solution. But still it doesn't solve the question. As they all seemed to be be true. Which part I am not getting correctly ? $\endgroup$ Commented Nov 11, 2021 at 2:15

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